Question

What is \[7^0 + \sqrt{7^1 + \sqrt{7^2 + \sqrt{7^4 + \sqrt{7^8 + \sqrt{7^{16} + \sqrt{7^{32} + \sqrt{7^{64} + ...}}}}}}}\] equal to?

Answer 1

Might as well generalize the problem now to any positive integer n

Answer 2

\[y = 7^0 + \sqrt{7^1 + \sqrt{7^2 + \sqrt{7^4 + \sqrt{7^8 + \sqrt{7^{16} + \sqrt{7^{32} + \sqrt{7^{64} + ...}}}}}}}\]
\[y = 1 + \sqrt{y}\]
maybe I am not sure

Answer 3

And I notice agdgd that you like to ask questions but I haven't seen you solve many yet.

So let's see something. Show that
\[1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + ...}}}}}}}\] converges.

Answer 4

@ishaan: see agdgd's earlier post.

Answer 5

\[n^0 + \sqrt{n^1 + \sqrt{n^2 + \sqrt{n^4+\sqrt{n^8+\sqrt{n^{16}+...}}}}}=1 + \sqrt{n}*\phi = \frac{\sqrt{5n} + \sqrt{n}+2}{2}\]

Answer 6

Oh Okay thanks

Answer 7

I'm not too good at solving questions :(

Answer 8

so we use induction?

Answer 9

oh well somebody else solved this problem already: http://www.math.washington.edu/~morrow/336_11/papers/brian.pdf

Answer 10

better than I ever could

Answer 11

so we do use induction :-D

Answer 12

hmm is it \[y = 1 + \sqrt{7y}\]

Answer 13

oh no sorry mistake my bad

Answer 14

so we define \[\sqrt{a+\sqrt{a+\sqrt{a+...}}}\]as \[t_n(a)\]where\[t_1(a) = \sqrt{a}\] and \[t_{n+1}(a)=\sqrt{a +t_n}\]

Answer 15

how does induction work here?

Answer 16

step 1: show that t(1) is convergent
that's easy enough

Answer 17

The paper you link to glosses the proof that it converges, just saying it's "easy".

Answer 18

I hope it is easy enough for me :(

Answer 19

so I think now I have to show that \[t_{n+1}(1)=\sqrt{1 +t_n}\] is convergent?

Answer 20

so we can say that as n approaches infinity, t_n+1 = t_n ?

Answer 21

\[t_n(1)^2 - t_n(1) -1 = 0\]

Answer 22

You show it's convergent by showing that the sequence (tn) is monotonically increasing and bounded above.

Answer 23

so I find the derivative?

Answer 24

Nope.

Answer 25

Show that \( t_{n+1} > t_n \)

Answer 26

using induction?

Answer 27

and show there exists a number B such that \( t_n < B \) for all \( n \).

Answer 28

Differentiate it

Answer 29

nope. It's not a continuous function.

Answer 30

Oh sorry

Answer 31

so t_0(1) = 1, t_1(1) = \sqrt(2), t_2(1) = \sqrt{1 + \sqrt{2}} then what?

Answer 32

1. First show that \( \varphi\) is an upper bound of the sequence by showing that
\[ x < \varphi \ \implies \sqrt{1+x} < \varphi \ \ \ --- (*)\]
If you prove that, then as we know that \( 1 < \varphi \), it will follow by the PMI that
\( t_n(1) < \varphi \) for all \( n \geq 1 \).

2. Then show that \( (t_n) \) is monotonically increasing. It will turn out the result of the 1. will be critical.

It is then time to use one of the most basic and important theorems in analysis:
If \( (x_n) \) is a monotonically increasing sequence of real numbers bounded above, then \( (x_n) \) converges. Don't try and prove this theorem; just use it. You can find the proof in Spivak or any other first book on analysis.

Answer 33

I think step 1 can be done by showing something about phi.

Answer 34

Try it for once instead of second guessing:

Suppose
\[ x < \varphi \]
then
\[ \sqrt{1+x} < \sqrt{1 + \varphi} = ... \]

Answer 35

it follows from \[\phi = \sqrt{\phi + 1}\]that \[\sqrt{1+x}<\phi\]

Answer 36

but then what do I do next?

Answer 37

so I simply plug in t_n?

Answer 38

Prove that
\[ \varphi = \sqrt{1 + \varphi} \]

Answer 39

http://en.wikipedia.org/wiki/Golden_ratio#Calculation

Answer 40

You show it.

Answer 41

how could I do that without knowing what phi is?

Answer 42

We're starting this chain of logic with
\[ \varphi = \frac{\sqrt{5}+1}{2} \]

Answer 43

I think thinking about golden ratios and all for this problem is not going to help, golden ratio like pi keeps popping in various things in mathematics like this one ..

Answer 44

:(

Answer 45

Actually, it's pretty much essential for showing that this sequence \( (t_n) \) converges.

Answer 46

I think agd problem is he/he is somehow trying to learn everything at once may be that's causing cognitive overflow ?

Answer 47

Maybe i simply don't get induction

Answer 48

or maybe i'm not putting in the effort

Answer 49

i'm too used to uncle Google and the Spivak answer book

Answer 50

@James there are several other approaches see here : http://math.stackexchange.com/questions/11945/limit-of-the-nested-radical-sqrt7-sqrt7-sqrt7-cdots/11969#11969

and

http://math.stackexchange.com/questions/61048/rigorous-definition-for-convergence-of-a-nested-radical-sqrta-1-sqrta-2

Answer 51

Let \[ \varphi = \frac{\sqrt{5}+1}{2} \] and define a sequence \( (t_n) \) by
\[ t_1 = 1, \ \ t_{n+1} = \sqrt{1 + t_n} \ \ \hbox{ for } n \geq 1\]

Let P(n) be the statement \( t_n < \varphi \).

P(1) is the statement \( 1 < \varphi \). This is clearly true as \( 2 < \sqrt{5} + 1 \) because \( 1 < \sqrt{5} \) because \( 1 < 5 \).

Now we want to show that for \( k \geq 1 \), the statement \( P(k) \) implies \( P(k+1) \).

Suppose P(k). I.e., \( t_k < \varphi \). Then
\( \ \ \ \ \ t_{n+1} \)
\[ = \sqrt{1 + t_n} \]
\[ < \sqrt{1 + \varphi} \ \ \ \hbox{by hypothesis, i.e., by the statement P(k)} \]

Now
\[ \varphi = \sqrt{1 + \varphi} \]
iff
\[ \varphi^2 = 1 + \varphi \]
iff
\[ \left( \frac{\sqrt{5}+1}{2} \right)^2 = 1 + \frac{\sqrt{5}+1}{2} \]
iff
\[ \frac{6 + 2\sqrt{5}}{4} = \frac{3 + \sqrt{5}}{2} \]
which is true.

Therefore P(k) => P(k+1) for an arbitrary \( k \geq 1 \).

Hence, as P(1) is also true, it follows by the Principle of Mathematical Induction that P(n) is true for all \( n \geq 1 \).

Answer 52

Now we have shown that \( (t_n) \) is bounded above. This was the statement P(n).

Now YOU show that \( (t_n) \) is an strictly increasing sequence.

Answer 53

Great answer James but may be you are expecting too much out of a guy who probably didn't studied real analysis or probably analysis before ? ;)

Answer 54

We need to show that \[t_{n+1} - t_n > 0 \text{ for all n}\geq1\]squaring both sides gives us \[(1 + t_n) - 2(t_{n+1} * t_n) + t_n^2 > 0\]

Answer 55

This will keep me up all day :(

Answer 56

or how about \[1 + t_n > t_n^2\]iff\[-(t^2 -t_n -1) > 0\]

Answer 57

t_n^2 * oops

Answer 58

\[-(t_n^2 -t_n -1) > 0\]iff \[t_n <\frac{1+\sqrt{5}}{2}\]this is true since \[t_{n+1} <\phi\]iff\[\sqrt{1 + t_n}<\sqrt{1 + \phi}\]iff\[t_n < \phi\]
so I think we've successfully shown that the sequence t_n is monotonically increasing

Answer 59

why is this step true?

Answer 60

if you find the roots of the polynomial, it turns out that \[\frac{1+\sqrt{5}}{2}\]is the upper bound

Answer 61

ok. You're not getting an A in analysis any time soon, but you're not getting an F today either. :-)

Answer 62

:( it took me an hour to figure that one out :-P

Answer 63

btw, when did you change into a black man? or were you always a black man?

Answer 64

I've always been george agdgdgwngo

Answer 65

then how come when I first knew you you had a picture of a young girl as your profile pick?

Answer 66

I've been a young girl once.

Answer 67

by which I mean 19 years old or so.

Answer 68

ok then.

Answer 69

How do I proceed from all this?

Answer 70

If I ever want to master the 'proof by induction'

Answer 71

like I told you when you first asked about induction, you need to write a few out, get them corrected, like just about everything in mathematics. Failing that, read carefully written induction proofs, like the one I've written in this thread, or in good books like Spivak.

Answer 72

If you want to practice, write out a proof for by induction for this identity:
1!1 + 2!2 + 3!3 + ... + n!n = (n+1)! - 1.

Write out the proof in four sections, beginning exactly as follows:

"Let \(P(n)\) be the statement ..."

"First, the statement \( P(1) \) is true because ..."

"Next we want to show that for \( k \geq 1 \ P(k)\implies P(k+1)\). \( P(k) \) says that ..."

"Hence having shown that \(P(1)\) is true and that for \( k \geq 1 \), \( P(k)\implies P(k+1) \), it follows by the Principle of Mathematical Induction that ..."

Answer 73

\[\small{\text{Let }}P(n)\small{\text{ be the following statement:}}\]\[1!1 + 2!2 + 3!3 + ... + n!n = (n+1)! - 1\]\[\small\text{Hence,}\] P(1)\[\text{ is the statement}\]\[1!1 = (1+1)!-1\]iff\[1=1\]which is true
\[\text{Next we want to show that for }k \geq1, \ P(k)\implies P(k+1).\text{}\]\[\text{If all of }P(1), P(2)...P(k)\text{ are true, then in particular, }P(k)\text{ is true.}\]\[P(k)\text{ says that:}\]\[1!1 + 2!2 + 3!3 + ... + k!k = (k+1)! - 1\]adding \[(k+1)!*(k+1)\]to both sides gives us \[1!1 + 2!2 + 3!3 +...+ k!k + (k+1)!(k+1) = (k+2)!-1\]which shoes that P(k+1) is also true.
Hence having shown that P(1) is true and that for k≥1, P(k)⟹P(k+1), it follows by the Principle of Mathematical Induction that P(n) is true for all n \geq 1

Answer 74

Am I mistaken?

Answer 75

Yes, four small and one large problems, most of which are the second step of the proof, "P(k) => P(k+1)"

* In the first step, it would be better to say that
1!1 = 2! - 1
iff
1 = 2 - 1

* We are NOT assuming all of P(1), P(2), ... P(k-1), P(k).
We are only using the hypothesis P(k).

* In this step we do not say P(k) is true, as we don't know that yet. We say
"suppose P(k)"; we are using that statement as a hypothesis.

*** This is the most serious issue:
In the heart of the proof, after you've added (k+1)!(k+1) to both sides, you
skipped the critical calculations in the RHS. Why is it true that
(k+1)! - 1 + (k+1)!(k+1) = (k+2)! - 1 ?

I'm not asking you write it out for me now; I'm simply saying this is the most
important part of the proof and you skipped over it.

* At the end of this step, we do not say "hence P(k+1) is true", as we still don't
know that yet. All we have shown is that P(k) => P(k+1). Remember that
implication can be true without P(k) being true.